I am trying to understand Remark 7.2.22 (Page 256) of Algebraic Function Fields and Codes (Second Edition) by Henning Stichtenoth.

In that remark he considers a tower $\mathcal{F} = (F_0,F_1,F_2,\dots)$ over a finite constant field $\mathbb{F_q}$ and extends it constantly with $L$ to get $\mathcal{F'}=(F_0',F_1',F_2',\dots)$ (i.e. $L/\mathbb{F_q}$ is an algebraic extension (may be finite or infinite) and $F_i' = F_iL$ $\forall i \geq 1$).

Next he takes a place

$P \in \mathbb{P}(F_i)$ for some $i >0$ and asserts that $P$ ramifies in the extension $F_{i+1}/F_i$ if and only if the places $P' \in \mathbb{P}(F_i')$ above $P$ are ramified in the extension $F_{i+1}'/F_i'$.

My doubt here is if the situation is as given below

Case 1: $$ \begin{array}{ccccccc} F_i' & \rightarrow & F_{i+1}' &&P'&\rightarrow & P_1'\\ \uparrow & & \uparrow && \uparrow && \uparrow\\ F_i& \rightarrow & F_{i+1} && P &\xrightarrow{e>1} & P_1 \end{array} $$ (where $e$ is the corresponding ramification index in the diagram) then as $e(P'/P)=e(P_1'/P_1)=1$ (because $F_i'/F_i$ is a constant field extension) we can say that $e(P_1'/P')=e(P_1/P) >1$.

But what if the situation as depicted in the following diagram also happens?

Case 2: $$ \begin{array}{ccc} Q & \rightarrow & P_2'\\ \uparrow && \uparrow \\ P & \xrightarrow{e=1} & P_2 \end{array} $$ i.e. $P_1, P_2$ both extend $P$ with one extension ramified and another unramified. In this case $e(P_2'|Q)$ has to be equal to $1$.

How can he conclude that every place $P' \in \mathbb{P}(F_i')$ gets ramified if $P$ is ramified? Doesn't Case 2 situation arise?

p.s. I have asked the same question in math.stackexchange.com but I could not get an answer there (update: now there's an answer).

  • $\begingroup$ I think the argument uses that $L/\mathbb{F}_q$ is Galois. Then every $F_i'/F_i$ is Galois and the extensions $F_{i+1}'/F_i'$ are Galois equivariant. Moreover, the Galois group acts transitively on the places of $F_i'$ lying over the place $P$ of $F_i$. Therefore, all the places over $P$ have the same splitting behaviour, if one is ramified then all are. $\endgroup$ Nov 3, 2015 at 15:11
  • $\begingroup$ If it is Galois then everything make sense but in the book $L/\mathbb{F}_q$ is not assumed to be Galois. $\endgroup$
    – Bharath
    Nov 3, 2015 at 15:42
  • $\begingroup$ But $L/\mathbb{F}_q$ is an algebraic extension of a finite field, so it is Galois. $\endgroup$ Nov 3, 2015 at 18:20
  • $\begingroup$ Yeah you are right. Thanks for clearing my doubt. $\endgroup$
    – Bharath
    Nov 3, 2015 at 18:48

1 Answer 1


The question was settled in the comments. The extension $L/\mathbb{F}_q$ is Galois, hence the extensions $F_i'/F_i$ are Galois and the extensions $F_{i+1}'/F_i'$ are equivariant for the Galois action. For any given place $P$ of $F_i$, the Galois group $\operatorname{Gal}(F_i'/F_i)$ acts transitively on the set of places $P'$ lying over $P$. This means that all places $P'$ over $P$ have the same splitting behaviour, i.e., same ramification and inertia degree. Hence, if one place of $F_i'$ ramifies in the extension $F_{i+1}'/F_i'$, then all places ramify.

  • $\begingroup$ I am sorry Wendt. I think I have jumped a bit to conclude. The problem is still there. We are not talking about $e(P'|P)$ because that is obviously unramified. We are talking about $e(P_1'|P')$. The places $P'$ lying above $P$ have same ramification index because $F_i'/F_i$ is Galois but we dont know about $F_{i+1}'/F_i'$. So we cannot say $e(P_2'|Q)=e(P_1'|P')=1$. (Check the notations from the above post). So I think one should assume that the extension $F_{i+1}/F_i$ is Galois. $\endgroup$
    – Bharath
    Nov 22, 2015 at 6:42

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